Circle geometry

2. Chord theorems

ID is: 3933 Seed is: 9952

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

$A, B$ , and $C$ are points on the circle having centre $O$ . $S$ and $T$ are points on $AC$ and $AB$ respectively such that $OS \perp AC$ and $OT \perp AB$ . $AB = 40$ and $AC = 48$ .

NOTE: Do not assume that the diagram is drawn to scale.

Calculate $AS$ .

Answer:
$AS =$

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the theorem about the perpendicular lines from the centre of a circle.

STEP: Apply the perpendicular line from the centre of a circle theorem
[−1 point ⇒ 0 / 1 points left]

We need to determine the length of $AS$ . We know that $AC$ has a length of $48$ . To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

The segment $OS$ is perpendicular to chord $AC$ . Therefore $OS$ bisects $AC$ . And that means $AS$ is half as long as $AC$ . We can use that fact to find the length of $AS$ .

$\begin{aligned} AS & = \frac{AC}{2} \\ & = \frac{48}{2} \\ & = 24 \end{aligned}$

$AS$ has a length of $24$ .

Submit your answer as:
If $OS = \dfrac{7}{15} OT$ , calculate the radius $OA$ of the circle.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer:
$OA =$

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.

STEP: Label the diagram with known information
[−1 point ⇒ 4 / 5 points left]

We need to find the radius of the circle (the length of $OA$ ). Let's start by labelling the information we know about the figure, and the radius that we want to find. In $\Delta OTA$ , side $TA$ must be half as long as $AB$ . This is for the same reason we used in Question 1. So $TA = 20$ , as labelled below. We are using $d$ for the length of $OT$ .

STEP: Write an equation for each of the right-angled triangles
[−1 point ⇒ 3 / 5 points left]

There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In $\Delta OSA$ :

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ r^{2} & = OS^{2} + SA^{2} \\ r^{2} & = \left(\dfrac{7}{15} d\right)^{2} + \left(24\right)^{2} \end{aligned}$

We can write a similar equation for $\Delta OTA$ .

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ r^{2} & = OT^{2} + TA^{2} \\ r^{2} & = \left(d\right)^{2} + \left(20\right)^{2} \end{aligned}$

STEP: Combine the equations and solve for $d$
[−2 points ⇒ 1 / 5 points left]

Now we have two equations that are both equal to $r^{2}$ . So they must be equal to each other, and we can write:

$\begin{aligned} d^{2} + 20^{2} & = \left(\dfrac{7}{15} d\right)^{2} + 24^{2} \end{aligned}$

NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

Now solve the equation for $d$ (which is the length of $OT$ ).

$\begin{aligned} d^{2} + 20^{2} & = \left(\dfrac{7}{15} \right)^{2} d^{2} + 24^{2} \\ d^{2} + 400 & = \dfrac{49}{225} d^{2} + 576 \\ d^{2} - \dfrac{49}{225} d^{2} & = 576 - 400 \\ \dfrac{176}{225} d^{2} & = 176 \\ d^{2} & = 225 \\ d & = \pm 15 \end{aligned}$

The result we want is a distance, so we can throw away the negative answer.

STEP: Calculate the length of the radius
[−1 point ⇒ 0 / 5 points left]

We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

$\begin{aligned} r^{2} & = \left( d \right)^{2} + \left(20\right)^{2} \\ r^{2} & = \left(15\right)^{2} + 400 \\ r^{2} & = 225 + 400 \\ & = 625 \\ \therefore r & = \pm 25 \end{aligned}$

Again, we throw away the negative answer because we are calculating a distance. The length of radius $OA$ is $25$ .

Submit your answer as:

ID is: 3933 Seed is: 6358

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

$A, B$ , and $C$ are points on the circle having centre $O$ . $S$ and $T$ are points on $AC$ and $AB$ respectively such that $OS \perp AC$ and $OT \perp AB$ . $AB = 136$ and $AC = 154$ .

NOTE: Do not assume that the diagram is drawn to scale.

Calculate $AS$ .

Answer:
$AS =$

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the theorem about the perpendicular lines from the centre of a circle.

STEP: Apply the perpendicular line from the centre of a circle theorem
[−1 point ⇒ 0 / 1 points left]

We need to determine the length of $AS$ . We know that $AC$ has a length of $154$ . To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

The segment $OS$ is perpendicular to chord $AC$ . Therefore $OS$ bisects $AC$ . And that means $AS$ is half as long as $AC$ . We can use that fact to find the length of $AS$ .

$\begin{aligned} AS & = \frac{AC}{2} \\ & = \frac{154}{2} \\ & = 77 \end{aligned}$

$AS$ has a length of $77$ .

Submit your answer as:
If $OS = \dfrac{12}{17} OT$ , calculate the radius $OA$ of the circle.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer:
$OA =$

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.

STEP: Label the diagram with known information
[−1 point ⇒ 4 / 5 points left]

We need to find the radius of the circle (the length of $OA$ ). Let's start by labelling the information we know about the figure, and the radius that we want to find. In $\Delta OTA$ , side $TA$ must be half as long as $AB$ . This is for the same reason we used in Question 1. So $TA = 68$ , as labelled below. We are using $d$ for the length of $OT$ .

STEP: Write an equation for each of the right-angled triangles
[−1 point ⇒ 3 / 5 points left]

There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In $\Delta OSA$ :

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ r^{2} & = OS^{2} + SA^{2} \\ r^{2} & = \left(\dfrac{12}{17} d\right)^{2} + \left(77\right)^{2} \end{aligned}$

We can write a similar equation for $\Delta OTA$ .

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ r^{2} & = OT^{2} + TA^{2} \\ r^{2} & = \left(d\right)^{2} + \left(68\right)^{2} \end{aligned}$

STEP: Combine the equations and solve for $d$
[−2 points ⇒ 1 / 5 points left]

Now we have two equations that are both equal to $r^{2}$ . So they must be equal to each other, and we can write:

$\begin{aligned} d^{2} + 68^{2} & = \left(\dfrac{12}{17} d\right)^{2} + 77^{2} \end{aligned}$

NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

Now solve the equation for $d$ (which is the length of $OT$ ).

$\begin{aligned} d^{2} + 68^{2} & = \left(\dfrac{12}{17} \right)^{2} d^{2} + 77^{2} \\ d^{2} + 4624 & = \dfrac{144}{289} d^{2} + 5929 \\ d^{2} - \dfrac{144}{289} d^{2} & = 5929 - 4624 \\ \dfrac{145}{289} d^{2} & = 1305 \\ d^{2} & = 2601 \\ d & = \pm 51 \end{aligned}$

The result we want is a distance, so we can throw away the negative answer.

STEP: Calculate the length of the radius
[−1 point ⇒ 0 / 5 points left]

We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

$\begin{aligned} r^{2} & = \left( d \right)^{2} + \left(68\right)^{2} \\ r^{2} & = \left(51\right)^{2} + 4624 \\ r^{2} & = 2601 + 4624 \\ & = 7225 \\ \therefore r & = \pm 85 \end{aligned}$

Again, we throw away the negative answer because we are calculating a distance. The length of radius $OA$ is $85$ .

Submit your answer as:

ID is: 3933 Seed is: 3131

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

$A, B$ , and $C$ are points on the circle having centre $O$ . $X$ and $Y$ are points on $AC$ and $AB$ respectively such that $OX \perp AC$ and $OY \perp AB$ . $AB = 40$ and $AC = 48$ .

NOTE: Do not assume that the diagram is drawn to scale.

Calculate $AY$ .

Answer:
$AY =$

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the theorem about the perpendicular lines from the centre of a circle.

STEP: Apply the perpendicular line from the centre of a circle theorem
[−1 point ⇒ 0 / 1 points left]

We need to determine the length of $AY$ . We know that $AB$ has a length of $40$ . To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

The segment $OY$ is perpendicular to chord $AB$ . Therefore $OY$ bisects $AB$ . And that means $AY$ is half as long as $AB$ . We can use that fact to find the length of $AY$ .

$\begin{aligned} AY & = \frac{AB}{2} \\ & = \frac{40}{2} \\ & = 20 \end{aligned}$

$AY$ has a length of $20$ .

Submit your answer as:
If $OX = \dfrac{7}{15} OY$ , calculate the radius $OA$ of the circle.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer:
$OA =$

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.

STEP: Label the diagram with known information
[−1 point ⇒ 4 / 5 points left]

We need to find the radius of the circle (the length of $OA$ ). Let's start by labelling the information we know about the figure, and the radius that we want to find. In $\Delta OXA$ , side $XA$ must be half as long as $AC$ . This is for the same reason we used in Question 1. So $XA = 24$ , as labelled below. We are using $d$ for the length of $OY$ .

STEP: Write an equation for each of the right-angled triangles
[−1 point ⇒ 3 / 5 points left]

There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In $\Delta OXA$ :

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ r^{2} & = OX^{2} + XA^{2} \\ r^{2} & = \left(\dfrac{7}{15} d\right)^{2} + \left(24\right)^{2} \end{aligned}$

We can write a similar equation for $\Delta OYA$ .

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ r^{2} & = OY^{2} + YA^{2} \\ r^{2} & = \left(d\right)^{2} + \left(20\right)^{2} \end{aligned}$

STEP: Combine the equations and solve for $d$
[−2 points ⇒ 1 / 5 points left]

Now we have two equations that are both equal to $r^{2}$ . So they must be equal to each other, and we can write:

$\begin{aligned} d^{2} + 20^{2} & = \left(\dfrac{7}{15} d\right)^{2} + 24^{2} \end{aligned}$

NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

Now solve the equation for $d$ (which is the length of $OY$ ).

$\begin{aligned} d^{2} + 20^{2} & = \left(\dfrac{7}{15} \right)^{2} d^{2} + 24^{2} \\ d^{2} + 400 & = \dfrac{49}{225} d^{2} + 576 \\ d^{2} - \dfrac{49}{225} d^{2} & = 576 - 400 \\ \dfrac{176}{225} d^{2} & = 176 \\ d^{2} & = 225 \\ d & = \pm 15 \end{aligned}$

The result we want is a distance, so we can throw away the negative answer.

STEP: Calculate the length of the radius
[−1 point ⇒ 0 / 5 points left]

We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

$\begin{aligned} r^{2} & = \left( d \right)^{2} + \left(20\right)^{2} \\ r^{2} & = \left(15\right)^{2} + 400 \\ r^{2} & = 225 + 400 \\ & = 625 \\ \therefore r & = \pm 25 \end{aligned}$

Again, we throw away the negative answer because we are calculating a distance. The length of radius $OA$ is $25$ .

Submit your answer as:

ID is: 1626 Seed is: 983

Using the perpendicular bisector theorem

In the circle with centre $O$ , segment $\overline{JK} = \text{24}$ . $\overline{OE}$ is perpendicular to $\overline{JK}$ at Point $E$ . This is shown in the figure below, which is drawn to scale.

Complete the sentence below by choosing the correct options.

Answer:
$\overline{EK}$ is because .
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'

STEP: Use the perpendicular bisector theorem
[−2 points ⇒ 0 / 2 points left]

One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

In this figure, $\overline{JK}$ is the chord, and it is bisected at point $E$ . In other words, $E$ is the midpoint of $\overline{JK}$ . We know the chord is bisected because it is perpendicular to segment $\overline{OE}$ .

On the diagram below the segments $\overline{JE}$ and $\overline{EK}$ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

Based on this, we can get the answers: $\overline{EK}$ is "half the length of $\overline{JK}$ " because " $\perp$ line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).

Submit your answer as: and
Now find the length of $\overline{EK}$ in the diagram.

Answer: $\overline{EK} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!

STEP: Use the answer we found in Question 1 to calculate the answer
[−1 point ⇒ 0 / 1 points left]

The length that we want is smaller than the length given. Therefore, we must divide by two to get the length of the smaller segment (this is the same as multiplying by a half).

$\begin{aligned} \text{smaller} & = \frac{1}{2} \times \text{bigger} \\ \overline{EK} & = \frac{1}{2} \left(\overline{JK} \right) \\ & = \frac{1}{2} \left( \text{24} \right) \\ & = \text{12} \end{aligned}$

The length of $\overline{EK}$ is $\text{12}$ .

Submit your answer as:

ID is: 1626 Seed is: 1826

Using the perpendicular bisector theorem

In the circle with centre $O$ , segment $\overline{FJ} = \text{6}$ . $\overline{OF}$ is perpendicular to $\overline{KJ}$ at Point $F$ . This is shown in the figure below, which is drawn to scale.

Complete the sentence below by choosing the correct options.

Answer:
$\overline{KJ}$ is because .
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'

STEP: Use the perpendicular bisector theorem
[−2 points ⇒ 0 / 2 points left]

One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

In this figure, $\overline{KJ}$ is the chord, and it is bisected at point $F$ . In other words, $F$ is the midpoint of $\overline{KJ}$ . We know the chord is bisected because it is perpendicular to segment $\overline{OF}$ .

On the diagram below the segments $\overline{KF}$ and $\overline{FJ}$ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

Based on this, we can get the answers: $\overline{KJ}$ is "twice as long as $\overline{FJ}$ " because " $\perp$ line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).

Submit your answer as: and
Now find the length of $\overline{KJ}$ in the diagram.

Answer: $\overline{KJ} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!

STEP: Use the answer we found in Question 1 to calculate the answer
[−1 point ⇒ 0 / 1 points left]

The length that we want is larger than the length given. Therefore, we must multiply by two to get the length of the chord.

$\begin{aligned} \text{larger} & = 2 \times \text{shorter} \\ \overline{KJ} & = 2 \left(\overline{FJ} \right) \\ & = 2 \left( \text{6} \right) \\ & = \text{12} \end{aligned}$

The length of $\overline{KJ}$ is $\text{12}$ .

Submit your answer as:

ID is: 1626 Seed is: 4981

Using the perpendicular bisector theorem

In the circle with centre $O$ , segment $\overline{KA} = \text{15}$ . $\overline{OA}$ is perpendicular to $\overline{KJ}$ at Point $A$ . This is shown in the figure below, which is drawn to scale.

Complete the sentence below by choosing the correct options.

Answer:
$\overline{KJ}$ is because .
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'

STEP: Use the perpendicular bisector theorem
[−2 points ⇒ 0 / 2 points left]

One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

In this figure, $\overline{KJ}$ is the chord, and it is bisected at point $A$ . In other words, $A$ is the midpoint of $\overline{KJ}$ . We know the chord is bisected because it is perpendicular to segment $\overline{OA}$ .

On the diagram below the segments $\overline{KA}$ and $\overline{AJ}$ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

Based on this, we can get the answers: $\overline{KJ}$ is "twice as long as $\overline{KA}$ " because " $\perp$ line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).

Submit your answer as: and
Now find the length of $\overline{KJ}$ in the diagram.

Answer: $\overline{KJ} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!

STEP: Use the answer we found in Question 1 to calculate the answer
[−1 point ⇒ 0 / 1 points left]

The length that we want is larger than the length given. Therefore, we must multiply by two to get the length of the chord.

$\begin{aligned} \text{larger} & = 2 \times \text{smaller} \\ \overline{KJ} & = 2 \left(\overline{KA} \right) \\ & = 2 \left( \text{15} \right) \\ & = \text{30} \end{aligned}$

The length of $\overline{KJ}$ is $\text{30}$ .

Submit your answer as:

ID is: 1625 Seed is: 5877

Calculations using the 'perpendicular bisector' theorem

In the circle $O$ you are given the length of the chord $\overline{AP} = \text{8}$ and $\overline{OP} = \text{6}$ . Also, $\overline{OB} \perp \overline{AP}$ at point $B$ . Compute the length of $\overline{OB}$ .

INSTRUCTION: Round your answer to two decimal places (if necessary).

Answer: The length of

\bar{O B}

$\overline{OB}$ is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord $\overline{AP}$ . This means you can use the perpendicular bisector theorem.

STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

the length of $\overline{AP} = \text{8}$
length of $\overline{OP} = \text{6}$ which is the radius of the circle
$\overline{OB}$ is perpendicular to $\overline{AP}$

and we want:

the length of the side $\overline{OB}$

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord $\overline{AP}$ , and there is a line from the centre of the circle $O$ perpendicular to the chord, meeting the chord at point $B$ . What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that $\overline{AB}$ and $\overline{BP}$ are half the size of $\overline{AP}$ , which is 8.

\begin{aligned} ∴ \bar{A B} = \bar{B P} & = \frac{\bar{A P}}{2} (⊥ line from centre bisects chord) \\ = \frac{(8)}{2} \\ = 4 \end{aligned}

$\begin{aligned} \therefore \overline{AB} = \overline{BP} & = \frac{\overline{AP}}{2} \quad (\perp \textrm{line from centre bisects chord}) \\ & = \frac{(8)}{2} \\ & = \text{4} \end{aligned}$

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that $\overline{OP}$ is the radius .

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment $\overline{OB}$ , the line segment you want to find, is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.

STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {\bar{O P}}^{2} & = {\bar{O B}}^{2} + {\bar{B P}}^{2} \\ (6)^{2} & = {\bar{O B}}^{2} + (4)^{2} \\ ∴ \bar{O B} & = \pm \sqrt{36 - 16} \\ = \pm \sqrt{20} \end{aligned}

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ \overline{OP}^2 & = \overline{OB}^2 + \overline{BP}^2 \\ (\text{6})^2 & = \overline{OB}^2 +(\text{4})^2 \\ \therefore \overline{OB} & = \pm \sqrt{\text{36} - \text{16}} \\ & = \pm \sqrt{\text{20}} \end{aligned}$

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: $\overline{OB} = \text{4.47}$ .

Submit your answer as:

ID is: 1625 Seed is: 7217

Calculations using the 'perpendicular bisector' theorem

In the circle $O$ you are given the length of the chord $\overline{KR} = \text{7}$ and $\overline{OC} = \text{5.5}$ . Also, $\overline{OC} \perp \overline{KR}$ at point $G$ . Compute the length of $\overline{OG}$ .

INSTRUCTION: Round your answer to two decimal places (if necessary).

Answer: The length of

\bar{O G}

$\overline{OG}$ is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord $\overline{KR}$ . This means you can use the perpendicular bisector theorem.

STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

the length of $\overline{KR} = \text{7}$
length of $\overline{OC} = \text{5.5}$ which is the radius of the circle
$\overline{OC}$ is perpendicular to $\overline{KR}$

and we want:

the length of the side $\overline{OG}$

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord $\overline{KR}$ , and there is a line from the centre of the circle $O$ perpendicular to the chord, meeting the chord at point $G$ . What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that $\overline{KG}$ and $\overline{GR}$ are half the size of $\overline{KR}$ , which is 7.

\begin{aligned} ∴ \bar{K G} = \bar{G R} & = \frac{\bar{K R}}{2} (⊥ line from centre bisects chord) \\ = \frac{(7)}{2} \\ = 3.5 \end{aligned}

$\begin{aligned} \therefore \overline{KG} = \overline{GR} & = \frac{\overline{KR}}{2} \quad (\perp \textrm{line from centre bisects chord}) \\ & = \frac{(7)}{2} \\ & = \text{3.5} \end{aligned}$

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that $\overline{OC}$ is the radius of the circle; and more importantly, it is equal to any other radius of the circle, e.g. $\overline{OC} \equiv \overline{OK}$ .

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment $\overline{OG}$ , the line segment you want to find, is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.

STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {\bar{O K}}^{2} & = {\bar{O G}}^{2} + {\bar{G K}}^{2} \\ (5.5)^{2} & = {\bar{O G}}^{2} + (3.5)^{2} \\ ∴ \bar{O G} & = \pm \sqrt{(30.25) - (12.25)} \\ = \pm \sqrt{18} \end{aligned}

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ \overline{OK}^2 & = \overline{OG}^2 + \overline{GK}^2 \\ (\text{5.5})^2 & = \overline{OG}^2 +(\text{3.5})^2 \\ \therefore \overline{OG} & = \pm \sqrt{(\text{30.25})-(\text{12.25})} \\ & = \pm \sqrt{\text{18}} \end{aligned}$

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: $\overline{OG} = \text{4.24}$ .

Submit your answer as:

ID is: 1625 Seed is: 4590

Calculations using the 'perpendicular bisector' theorem

In the circle $O$ you are given the length of the chord $\overline{DC} = \text{7}$ and $\overline{OC} = \text{4.5}$ . Also, $\overline{OR} \perp \overline{DC}$ at point $R$ . Compute the length of $\overline{OR}$ .

INSTRUCTION: Round your answer to two decimal places (if necessary).

Answer: The length of

\bar{O R}

$\overline{OR}$ is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord $\overline{DC}$ . This means you can use the perpendicular bisector theorem.

STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

the length of $\overline{DC} = \text{7}$
length of $\overline{OC} = \text{4.5}$ which is the radius of the circle
$\overline{OR}$ is perpendicular to $\overline{DC}$

and we want:

the length of the side $\overline{OR}$

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord $\overline{DC}$ , and there is a line from the centre of the circle $O$ perpendicular to the chord, meeting the chord at point $R$ . What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that $\overline{DR}$ and $\overline{RC}$ are half the size of $\overline{DC}$ , which is 7.

\begin{aligned} ∴ \bar{D R} = \bar{R C} & = \frac{\bar{D C}}{2} (⊥ line from centre bisects chord) \\ = \frac{(7)}{2} \\ = 3.5 \end{aligned}

$\begin{aligned} \therefore \overline{DR} = \overline{RC} & = \frac{\overline{DC}}{2} \quad (\perp \textrm{line from centre bisects chord}) \\ & = \frac{(7)}{2} \\ & = \text{3.5} \end{aligned}$

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that $\overline{OC}$ is the radius .

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment $\overline{OR}$ , the line segment you want to find, is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.

STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {\bar{O C}}^{2} & = {\bar{O R}}^{2} + {\bar{R C}}^{2} \\ (4.5)^{2} & = {\bar{O R}}^{2} + (3.5)^{2} \\ ∴ \bar{O R} & = \pm \sqrt{20.25 - 12.25} \\ = \pm \sqrt{8} \end{aligned}

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ \overline{OC}^2 & = \overline{OR}^2 + \overline{RC}^2 \\ (\text{4.5})^2 & = \overline{OR}^2 +(\text{3.5})^2 \\ \therefore \overline{OR} & = \pm \sqrt{\text{20.25} - \text{12.25}} \\ & = \pm \sqrt{\text{8}} \end{aligned}$

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: $\overline{OR} = \text{2.83}$ .

Submit your answer as:

ID is: 1627 Seed is: 9249

Using circle theorems to relate angles

In the circle with midpoint $M$ the arc $QA$ subtends the angles $x$ and $y$ . What is angle $y$ in terms of $x$ ? Please choose a reason for your answer from the options given.

Answer:

y =

$y =$ because:

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The words "in terms of

x

$x$ " mean that the answer will include

x

$x$ - it will not be a number alone. For example,

2 x^{3} - x

$2x^{3} - x$ and

5 x - 1

$5x- 1$ are "in terms of

x

$x$ ."

STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

We recall the circle theorem which states that angles subtended by arcs of equal length, or by the same arc, are equal. Here the arc is $QA$ and the angles $x$ and $y$ are both subtended by it. Therefore, the angles must be the same size: angle $y$ in terms of $x$ is $y = x$ .

The reason for this theorem is: $\angle\text{s subtended by same arc }QA$ .

Submit your answer as: and

ID is: 1627 Seed is: 2819

Using circle theorems to relate angles

In the circle with midpoint $M$ the arc $CJ$ subtends the angles $x$ and $y$ . What is angle $y$ in terms of $x$ ? Please choose a reason for your answer from the options given.

Answer:

y =

$y =$ because:

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The words "in terms of

x

$x$ " mean that the answer will include

x

$x$ - it will not be a number alone. For example,

2 x^{3} - x

$2x^{3} - x$ and

5 x - 1

$5x- 1$ are "in terms of

x

$x$ ."

STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

A useful circle theorem for this question is the following: the angle subtended by an arc at the centre of a circle is twice the size of the angle subtended by the same arc at the circumference of the circle. Here the arc is $CJ$ ; the angle subtended by the arc at the centre of a circle is $x$ ; and the angle subtended by the same arc at the circumference of the circle is $y$ . Therefore, $x$ is twice the size of $y$ or: $x = 2y$ . Writing angle $y$ in terms of $x$ , we get $y = \frac{1}{2}x$ .

The reason for this theorem is: $\angle \text{ at centre } = 2 \angle\text{ at circum.}$

Submit your answer as: and

ID is: 1627 Seed is: 727

Using circle theorems to relate angles

In the circle with midpoint $M$ the arc $KB$ subtends the angles $x$ and $y$ . What is angle $y$ in terms of $x$ ? Please choose a reason for your answer from the options given.

Answer:

y =

$y =$ because:

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The words "in terms of

x

$x$ " mean that the answer will include

x

$x$ - it will not be a number alone. For example,

2 x^{3} - x

$2x^{3} - x$ and

5 x - 1

$5x- 1$ are "in terms of

x

$x$ ."

STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

We recall the circle theorem which states that angles subtended by arcs of equal length, or by the same arc, are equal. Here the arc is $KB$ and the angles $x$ and $y$ are both subtended by it. Therefore, the angles must be the same size: angle $y$ in terms of $x$ is $y = x$ .

The reason for this theorem is: $\angle\text{s subtended by same arc }KB$ .

Submit your answer as: and

ID is: 4364 Seed is: 1064

Proof: angle at centre

Adapted from DBE Nov 2015 Grade 11, P2, Q10.1
Maths formulas

In the diagram, $O$ is the centre of the circle. $J$ , $K$ and $L$ are points on the circumference of the circle. $KO$ is produced to $M$ . $\hat{L} = y$ and $\hat{J} = x$ .

Write an expression for $\hat{O}_{1}$ in terms of $y$ .

Answer:
$OL$ and $OK$ are radii.
$\therefore$ $= y$ ( $\angle$ s opp equal sides)
$\therefore \angle O_{1} =$

expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Radii are always the same length. $\Delta LOK$ has two sides that are radii.

STEP: Two sides in $\Delta LOK$ are radii
[−1 point ⇒ 2 / 3 points left]

$\begin{aligned} OL &= OK \text{ (radii)} \\ \therefore \hat{K}_{1} &= y\, \left(\angle \text{s opp equal sides}\right) \end{aligned}$

STEP: $\hat{O}_{1}$ is an exterior angle of $\Delta LOK$
[−2 points ⇒ 0 / 3 points left]

The exterior angle of a triangle is equal to the sum of the opposite interior angles (ext $\angle$ of $\Delta$ ).

$\begin{aligned} \therefore \hat{O}_{1} &= y + y\, \left(\text{ext } \angle \text{ of } \Delta\right) \\ &= 2y \end{aligned}$

Submit your answer as: andand
Hence, prove the theorem that states that the angle subtended by an arc at the centre is equal to twice the angle subtended by the same arc at the circumference. In other words, prove the theorem that states $\text{reflex } L\hat{O}J = 2L\hat{K}J$ .

Here's the diagram again, for your convenience:

Answer:
In the same way as in Question 1, $\angle O_{2} = 2x.$

$\text{reflex } \angle LOJ$
$\quad =$
$\quad =$
$\quad =$
$\quad = 2\angle LKJ$
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

$\begin{array}{c} \text{reflex } L\hat{O}J = \hat{O}_{1} + \hat{O}_{2} \end{array}$

Use the expressions that you have for $\hat{O}_{1}$ and $\hat{O}_{2}$ .

STEP: Use the expressions for $\hat{O}_{1}$ and $\hat{O}_{2}$ , then take out the common factor
[−2 points ⇒ 1 / 3 points left]

We are trying to prove that reflex ${\color{blue}{L\hat{O}J}}$ is double the size of ${\color{red}{L\hat{K}J}}$ . We are going to split up both angles into two parts and work with the sum.

In the same way as in Question 1, $\hat{O}_{2} = 2x$ .

$\begin{aligned} \therefore \text{reflex } L\hat{O}J &= 2y + 2x \\ &= 2(y + x) \end{aligned}$

STEP: Replace $y$ with $\hat{K}_{1}$ and $x$ with $\hat{K}_{2}$ , since they are equal
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \text{reflex } L\hat{O}J &= 2\left(\hat{K}_{1} + \hat{K}_{2}\right) \\ &= 2L\hat{K}J \end{aligned}$

Submit your answer as: andand

ID is: 4364 Seed is: 4561

Proof: angle at centre

Adapted from DBE Nov 2015 Grade 11, P2, Q10.1
Maths formulas

In the diagram, $O$ is the centre of the circle. $J$ , $K$ and $L$ are points on the circumference of the circle. $KO$ is produced to $M$ . $\hat{L} = y$ and $\hat{J} = x$ .

Write an expression for $\hat{O}_{2}$ in terms of $x$ .

Answer:
$OJ$ and $OK$ are radii.
$\therefore$ $= x$ ( $\angle$ s opp equal sides)
$\therefore \angle O_{2} =$

expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Radii are always the same length. $\Delta JOK$ has two sides that are radii.

STEP: Two sides in $\Delta JOK$ are radii
[−1 point ⇒ 2 / 3 points left]

$\begin{aligned} OJ &= OK \text{ (radii)} \\ \therefore \hat{K}_{2} &= x\, \left(\angle \text{s opp equal sides}\right) \end{aligned}$

STEP: $\hat{O}_{2}$ is an exterior angle of $\Delta JOK$
[−2 points ⇒ 0 / 3 points left]

The exterior angle of a triangle is equal to the sum of the opposite interior angles (ext $\angle$ of $\Delta$ ).

$\begin{aligned} \therefore \hat{O}_{2} &= x + x\, \left(\text{ext } \angle \text{ of } \Delta\right) \\ &= 2x \end{aligned}$

Submit your answer as: andand
Hence, prove the theorem that states that the angle subtended by an arc at the centre is equal to twice the angle subtended by the same arc at the circumference. In other words, prove the theorem that states $L\hat{O}J = 2L\hat{K}J$ .

Here's the diagram again, for your convenience:

Answer:
In the same way as in Question 1, $\angle O_{1} = 2y.$

$\angle LOJ$
$\quad =$
$\quad =$
$\quad =$
$\quad = 2\angle LKJ$
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

$\begin{array}{c} \text{ } L\hat{O}J = \hat{O}_{1} + \hat{O}_{2} \end{array}$

Use the expressions that you have for $\hat{O}_{1}$ and $\hat{O}_{2}$ .

STEP: Use the expressions for $\hat{O}_{1}$ and $\hat{O}_{2}$ , then take out the common factor
[−2 points ⇒ 1 / 3 points left]

We are trying to prove that ${\color{blue}{L\hat{O}J}}$ is double the size of ${\color{red}{L\hat{K}J}}$ . We are going to split up both angles into two parts and work with the sum.

In the same way as in Question 1, $\hat{O}_{1} = 2y$ .

$\begin{aligned} \therefore L\hat{O}J &= 2x + 2y \\ &= 2(x + y) \end{aligned}$

STEP: Replace $x$ with $\hat{K}_{2}$ and $y$ with $\hat{K}_{1}$ , since they are equal
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} L\hat{O}J &= 2\left(\hat{K}_{2} + \hat{K}_{1}\right) \\ &= 2L\hat{K}J \end{aligned}$

Submit your answer as: andand

ID is: 4364 Seed is: 2957

Proof: angle at centre

Adapted from DBE Nov 2015 Grade 11, P2, Q10.1
Maths formulas

In the diagram, $O$ is the centre of the circle. $A$ , $B$ and $C$ are points on the circumference of the circle. $CO$ is produced to $D$ . $\hat{B} = y$ and $\hat{A} = x$ .

Write an expression for $\hat{O}_{1}$ in terms of $y$ .

Answer:
$OB$ and $OC$ are radii.
$\therefore$ $\hat{C}_{1} = y$
$\therefore \hat{O}_{1} =$

expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Radii are always the same length. $\Delta BOC$ has two sides that are radii.

STEP: Two sides in $\Delta BOC$ are radii
[−1 point ⇒ 2 / 3 points left]

$\begin{aligned} OB &= OC \text{ (radii)} \\ \therefore \hat{C}_{1} &= y\, \left(\angle \text{s opp equal sides}\right) \end{aligned}$

STEP: $\hat{O}_{1}$ is an exterior angle of $\Delta BOC$
[−2 points ⇒ 0 / 3 points left]

The exterior angle of a triangle is equal to the sum of the opposite interior angles (ext $\angle$ of $\Delta$ ).

$\begin{aligned} \therefore \hat{O}_{1} &= y + y\, \left(\text{ext } \angle \text{ of } \Delta\right) \\ &= 2y \end{aligned}$

Submit your answer as: andand
Hence, prove the theorem that states that the angle subtended by an arc at the centre is equal to twice the angle subtended by the same arc at the circumference. In other words, prove the theorem that states $\text{reflex } B\hat{O}A = 2B\hat{C}A$ .

Here's the diagram again, for your convenience:

Answer:
In the same way as in Question 1, $\angle O_{2} = 2x.$

$\text{reflex } \angle BOA$
$\quad =$
$\quad =$
$\quad =$
$\quad = 2\angle BCA$
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

$\begin{array}{c} \text{reflex } B\hat{O}A = \hat{O}_{1} + \hat{O}_{2} \end{array}$

Use the expressions that you have for $\hat{O}_{1}$ and $\hat{O}_{2}$ .

STEP: Use the expressions for $\hat{O}_{1}$ and $\hat{O}_{2}$ , then take out the common factor
[−2 points ⇒ 1 / 3 points left]

We are trying to prove that reflex ${\color{blue}{B\hat{O}A}}$ is double the size of ${\color{red}{B\hat{C}A}}$ . We are going to split up both angles into two parts and work with the sum.

In the same way as in Question 1, $\hat{O}_{2} = 2x$ .

$\begin{aligned} \therefore \text{reflex } B\hat{O}A &= 2x + 2y \\ &= 2(x + y) \end{aligned}$

STEP: Replace $x$ with $\hat{C}_{2}$ and $y$ with $\hat{C}_{1}$ , since they are equal
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \text{reflex } B\hat{O}A &= 2\left(\hat{C}_{2} + \hat{C}_{1}\right) \\ &= 2B\hat{C}A \end{aligned}$

Submit your answer as: andand

ID is: 4386 Seed is: 831

Butterfly angles and friends

Adapted from DBE Nov 2015, Grade 11, P2, Q9
Maths formulas

In the diagram, $O$ is the centre of the circle. Diameter $BD$ subtends $B\hat{A}D$ at the circumference of the circle. $C$ is another point on the circumference and chords $BC$ and $AC$ are drawn such that $AC$ intersects $BD$ at $O$ . $\hat{C} = \text{31}\text{°}$ .

Calculate, giving reasons, the size of:

Answer:

$B\hat{A}D =$ $\text{°}$
$\hat{D} =$ $\text{°}$
$B\hat{O}A =$ $\text{°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of diameters, chords and triangles!

STEP: Find

B \hat{A} D

$B\hat{A}D$ using diameter

B D

$BD$
[−2 points ⇒ 4 / 6 points left]

We were given that $BD$ is a diameter. It subtends the angle $B\hat{A}D$ at the circumference.

$\therefore B\hat{A}D = \text{90}\text{°}$ ( $\angle$ in semi circle)

Choose the right reason!

✔ When we know that a chord is a diameter, we use $\angle$ in semi circle to prove that an angle is $\text{90}\text{°}$ .
✘ When we know that an angle is $\text{90}\text{°}$ , we use chord subtends $90°$ to prove that the chord is a diameter.

STEP: Find

\hat{D}

$\hat{D}$ using angles in the same segment
[−2 points ⇒ 2 / 6 points left]

The angles $\hat{C}$ and $\hat{D}$ are both subtended by the chord $BA$ at the circumference.

$\therefore \hat{D} = \hat{C} = \text{31}\text{°}$ ( $\angle$ s in the same seg)

STEP: Find

B \hat{O} A

$B\hat{O}A$ using angle at centre =

2 \times

$2 \times$ angle at circumference
[−2 points ⇒ 0 / 6 points left]

The angles $B\hat{O}A$ and $\hat{C}$ are both subtended by chord $BA$ . $B\hat{O}A$ is at the centre and $\hat{C}$ is at the circumference.

$\therefore 2 \times \text{31}\text{°} = B\hat{O}A$ ( $\angle$ at centre $= 2 \times \angle$ at circumference)

$\therefore B\hat{O}A = \text{62}\text{°}$

Submit your answer as: andandandandand

ID is: 4386 Seed is: 2517

Butterfly angles and friends

Adapted from DBE Nov 2015, Grade 11, P2, Q9
Maths formulas

In the diagram, $O$ is the centre of the circle. Diameter $FH$ subtends $F\hat{E}H$ at the circumference of the circle. $G$ is another point on the circumference and chords $FG$ and $EG$ are drawn such that $EG$ intersects $FH$ at $O$ . $\hat{G} = \text{53}\text{°}$ .

Calculate, giving reasons, the size of:

Answer:

$F\hat{E}H =$ $\text{°}$
$F\hat{O}E =$ $\text{°}$
$\hat{F}_{2} =$ $\text{°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of diameters, chords and triangles!

STEP: Find

F \hat{E} H

$F\hat{E}H$ using diameter

F H

$FH$
[−2 points ⇒ 4 / 6 points left]

We were given that $FH$ is a diameter. It subtends the angle $F\hat{E}H$ at the circumference.

$\therefore F\hat{E}H = \text{90}\text{°}$ ( $\angle$ in semi circle)

Choose the right reason!

✔ When we know that a chord is a diameter, we use $\angle$ in semi circle to prove that an angle is $\text{90}\text{°}$ .
✘ When we know that an angle is $\text{90}\text{°}$ , we use chord subtends $90°$ to prove that the chord is a diameter.

STEP: Find

F \hat{O} E

$F\hat{O}E$ using angle at centre =

2 \times

$2 \times$ angle at circumference
[−2 points ⇒ 2 / 6 points left]

The angles $F\hat{O}E$ and $\hat{G}$ are both subtended by chord $FE$ . $F\hat{O}E$ is at the centre and $\hat{G}$ is at the circumference.

$\therefore 2 \times \text{53}\text{°} = F\hat{O}E$ ( $\angle$ at centre $= 2 \times \angle$ at circumference)

$\therefore F\hat{O}E = \text{106}\text{°}$

STEP: Find

{\hat{F}}_{2}

$\hat{F}_{2}$ using

Δ F O G

$\Delta FOG$
[−2 points ⇒ 0 / 6 points left]

$OF = OG$ (radii)

$\therefore \hat{F}_{2} = \text{53}\text{°}$ ( $\angle$ s opp equal sides)

NOTE: As usual in Geometry, there is more than one correct way of finding

{\hat{F}}_{2}

$\hat{F}_{2}$ , but we can only show you one way here. If you justify your answer using correct reasons, you will still be marked correct.

Submit your answer as: andandandandand

ID is: 4386 Seed is: 3180

Butterfly angles and friends

Adapted from DBE Nov 2015, Grade 11, P2, Q9
Maths formulas

In the diagram, $O$ is the centre of the circle. Diameter $QS$ subtends $Q\hat{P}S$ at the circumference of the circle. $R$ is another point on the circumference and chords $QR$ and $PR$ are drawn. Radius $PO$ is drawn. $Q\hat{O}P = \text{80}\text{°}$ .

Calculate, giving reasons, the size of:

Answer:

$Q\hat{P}S =$ $\text{°}$
$\hat{S} =$ $\text{°}$
$\hat{Q}_{1} =$ $\text{°}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Use the properties of diameters, chords and triangles!

STEP: Find

Q \hat{P} S

$Q\hat{P}S$ using diameter

Q S

$QS$
[−2 points ⇒ 4 / 6 points left]

We were given that $QS$ is a diameter. It subtends the angle $Q\hat{P}S$ at the circumference.

$\therefore Q\hat{P}S = \text{90}\text{°}$ ( $\angle$ in semi circle)

Choose the right reason!

✔ When we know that a chord is a diameter, we use $\angle$ in semi circle to prove that an angle is $\text{90}\text{°}$ .
✘ When we know that an angle is $\text{90}\text{°}$ , we use chord subtends $90°$ to prove that the chord is a diameter.

STEP: Find

\hat{S}

$\hat{S}$ using angle at centre =

2 \times

$2 \times$ angle at circumference
[−2 points ⇒ 2 / 6 points left]

The angles $Q\hat{O}P$ and $\hat{S}$ are both subtended by chord $QP$ . $Q\hat{O}P$ is at the centre and $\hat{S}$ is at the circumference.

$\therefore 2 \times \hat{S} = \text{80}\text{°}$ ( $\angle$ at centre $= 2 \times \angle$ at circumference)

$\therefore \hat{S} = \text{40}\text{°}$

STEP: Find

{\hat{Q}}_{1}

$\hat{Q}_{1}$ using

Δ P Q S

$\Delta PQS$
[−2 points ⇒ 0 / 6 points left]

$\hat{Q}_{1} = \text{180}\text{°} - \text{90}\text{°} - \text{40}\text{°}$ (sum of $\angle$ s in $\Delta$ )

$\therefore \hat{Q}_{1} = \text{50}\text{°}$

NOTE: As usual in Geometry, there is more than one correct way of finding

{\hat{Q}}_{1}

$\hat{Q}_{1}$ , but we can only show you one way here. If you justify your answer using correct reasons, you will still be marked correct.

Submit your answer as: andandandandand

Column A	Column B
Tangent
Chord (non-diameter)
Arc

Column A	Column B
Tangent	F
Chord (non-diameter)	B
Arc	A

Column A	Column B
Diameter
Radius
Segment

Column A	Column B
Diameter	D
Radius	C
Segment	E

Column A	Column B
Arc
Radius
Tangent

Column A	Column B
A line that makes contact with a circle at one point on the circle.
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle.
The part of the circle that is cut off by a chord.

Column A	Column B
A part of the circumference of a circle.
The part of the circle that is cut off by a chord.
Any straight line segment from the centre of the circle to a point on the circumference.

Column A	Column B
A straight line segment joining the ends of an arc.
A line that makes contact with a circle at one point on the circle.
A part of the circumference of a circle.

Statement	Reason
	$\angle$ at centre = $2 \angle$ at circum.
$\angle O_{2} = 2 \angle Y$	$\angle$ at centre = $2 \angle$ at circum.
	$\angle$ s in a rev.
$\therefore$
$\therefore \angle W + \angle Y = 180°$

Statement	Reason
$\angle O_{1} = 2\angle W$	$\angle$ at centre = $2 \angle$ at circum.
$\angle O_{2} = 2\angle Y$	$\angle$ at centre = $2 \angle$ at circum.
$\angle O_{1} + \angle O_{2} = 360°$	$\angle$ s in a rev.
$\therefore 2\angle W + 2\angle Y = 360°$
$\therefore \angle W + \angle Y = 180°$

Statement	Reason
$\angle O_{1} = 2 \angle A$	$\angle$ at centre = $2 \angle$ at circum.
	$\angle$ at centre = $2 \angle$ at circum.
	$\angle$ s in a rev.
$\therefore$
$\therefore \angle A + \angle C = 180°$

Statement	Reason
$\angle O_{1} = 2\angle A$	$\angle$ at centre = $2 \angle$ at circum.
$\angle O_{2} = 2\angle C$	$\angle$ at centre = $2 \angle$ at circum.
$\angle O_{1} + \angle O_{2} = 360°$	$\angle$ s in a rev.
$\therefore 2\angle A + 2\angle C = 360°$
$\therefore \angle A + \angle C = 180°$

Statement	Reason
$\angle O_{1} = 2 \angle W$	$\angle$ at centre = $2 \angle$ at circum.
	$\angle$ at centre = $2 \angle$ at circum.
	$\angle$ s in a rev.
$\therefore$
$\therefore \angle W + \angle Y = 180°$

✘	Construct $OV$ and $OY$ .
This is an incorrect construction.
✘	Construct radii.
It does not say which radii to draw.
✘	Construct $BY$ and $BV$ .
It does not say that $BY$ has to be a diameter.
✔	Construct radii $OY$ and $OW$ .

Statement	Reason
$\angle Y_2 + \angle Y_1 = 90\text{°}$
$\therefore \angle Y_2 =$
$OY = OW$
$\therefore$ $\angle W_2 = x$
$\therefore$ $\angle YOW =$ $180\text{°} - 2x$	(sum of $\angle$ s in $\Delta$ )
$\therefore \angle V =$	( $\angle$ at centre $= 2 \times \angle$ at circumference)

Statement	Reason
$FA = CA$
$\angle F_{1} = \angle C_{3}$
$\therefore$
$\therefore$ $TFAC$ is a cyclic quad

✘	Construct $VR$ and $VP$ .
It does not say that $VR$ has to be a diameter.
✔	Construct diameter $VOR$ . Join $V$ to $P$ .

✘	Construct diameter.
It does not say which diameter to construct, or that it must be joined up with $P$ .
✘	Construct $OP$ and $OR$ .
This is an incorrect construction.

Statement	Reason
$\angle R_1 + \angle R_2 = 90 \text{°}$
$\therefore \angle R_2 =$
$\angle P_1 = x$
$\angle P_1 + \angle P_2= 90\text{°}$
$\therefore$ $\angle P_2 =$

✘	Construct $OD$ and $OF$ .
This is an incorrect construction.
✘	Construct $CF$ and $CD$ .
It does not say that $CF$ has to be a diameter.
✘	Construct radii.
It does not say which radii to draw.
✔	Construct radii $OF$ and $OE$ .

Statement	Reason
	(tan $\bot$ radius)
$\therefore \angle F_2 =$ $90\text{°} - x$
$OF = OE$
$\therefore$ $\angle E_2 = x$
$\therefore$ $\angle FOE =$ $180\text{°} - 2x$
$\therefore \angle D =$	( $\angle$ at centre $= 2 \times \angle$ at circumference)

Statement	Reason
$\angle F_2 + \angle F_1 = 90\text{°}$	(tan $\bot$ radius)
$\therefore \angle F_2 = 90\text{°} - x$
$OF = OE$	(radii)
$\therefore \angle E_2 = x$	( $\angle$ s opp equal sides)
$\therefore \angle FOE = 180\text{°} - 2x$	(sum of $\angle$ s in $\Delta$ )
$\therefore \angle D =$ $90\text{°} - x$	( $\angle$ at centre $= 2 \times \angle$ at circumference)

Statement	Reason
$\angle C_{1} =$
Also,
$\therefore \angle C_{1} = \angle T_{3}$

Statement	Reason
$CB = FB$
	( $\angle$ s opp equal sides)
$\therefore$
$\therefore$ $TCBF$ is a cyclic quad

Statement	Reason
$\angle T_{2} =$
But, $\angle T_{3} =$
$\therefore \angle T_{2} = \angle T_{3}$
Therefore, $BT$ bisects $\angle CTF$ .

Statement	Reason

$\therefore \angle E_{3} = \angle F_{1}$	( $\angle$ s opp equal sides)
Also,
$\therefore \angle E_{3} = \angle T_{2}$

Statement	Reason
$\angle B = \angle E_{3}$
Also, $\angle E_{3} = \angle T_{3}$
$\therefore$
$\therefore$ $TC \parallel BE$

Circle geometry

1. Parts of circles

Circle geometry terminology

Circle geometry terminology

Circle geometry terminology

Circle geometry terminology

Circle geometry terminology

Circle geometry terminology

Vocabulary: subtends

Vocabulary: subtends

Vocabulary: subtends

2. Chord theorems

Using circle geometry to find the radius

Using circle geometry to find the radius

Using circle geometry to find the radius

Using the perpendicular bisector theorem

Using the perpendicular bisector theorem

Using the perpendicular bisector theorem

Calculations using the 'perpendicular bisector' theorem

Calculations using the 'perpendicular bisector' theorem

Calculations using the 'perpendicular bisector' theorem

Using circle theorems to relate angles

Using circle theorems to relate angles

Using circle theorems to relate angles

Proof: angle at centre

Proof: angle at centre

Proof: angle at centre

Butterfly angles and friends

Butterfly angles and friends

Butterfly angles and friends

3. Cyclic quadrilateral theorems

Opposite angles of a cyclic quadrilateral: Theorem

Opposite angles of a cyclic quadrilateral: Theorem

Opposite angles of a cyclic quadrilateral: Theorem

Circle theorems

Circle theorems

Circle theorems

What do cyclic quadrilaterals look like?

What do cyclic quadrilaterals look like?

What do cyclic quadrilaterals look like?

Overlapping circles: find equal angles

Overlapping circles: find equal angles

Overlapping circles: find equal angles

Overlapping circles: algebraic questions

Overlapping circles: algebraic questions

Overlapping circles: algebraic questions

Cyclic quads and parallel lines: algebraic questions

Cyclic quads and parallel lines: algebraic questions

Cyclic quads and parallel lines: algebraic questions

4. Tangent theorems

The tangent-chord theorem: finding congruent angles

The tangent-chord theorem: finding congruent angles

The tangent-chord theorem: finding congruent angles

Applying theorems from Euclidean geometry

Applying theorems from Euclidean geometry

Applying theorems from Euclidean geometry

Proving the tan chord theorem

Proving the tan chord theorem

Proving the tan chord theorem

Tangents and radii

Tangents and radii

Tangents and radii

The tangent-chord theorem: messy situations

The tangent-chord theorem: messy situations

The tangent-chord theorem: messy situations

Triangles and circles

Triangles and circles

Triangles and circles

Using the tangent-chord theorem to find angles

Using the tangent-chord theorem to find angles

Using the tangent-chord theorem to find angles

Two tangents from a single point

Two tangents from a single point

Two tangents from a single point

What do congruent tangent lines look like?

What do congruent tangent lines look like?

What do congruent tangent lines look like?

5. Mixed applications

Proving parallel lines using right angles

Proving parallel lines using right angles

Statement	Reason
$\angle C_{1} = \angle F_{2}$
Similarly,
But $TF = TE$
$\therefore \angle F_{2} = \angle E_{2}$
$\therefore \angle C_{1} = \angle C_{2}$
$\therefore TC$ bisects $\angle FCE$

Statement	Reason
$\angle F_{1} =$
Also,
$\therefore \angle F_{1} = \angle T_{3}$

Statement	Reason
$\therefore$ $\angle C_1$ $= 90\text{°} - x$	(sum of $\angle$ s in $\Delta$ )
$\therefore$ $= 90\text{°} - x$
$\angle OAD = 90\text{°}$
$\therefore$ $= x$
$\therefore$
$\therefore AOCD$ is a cyclic quadrilateral

Statement	Reason
$\angle F_{2} = \angle B_{1}$
$\therefore$ $= \text{90}\text{°}$
$\therefore$ $= \text{90}\text{°}$

A ✘	The circle has two diameters.
Every circle has infinitely many diameters. This sentence does not add anything to the proof.
B ✘	$TC \perp CB$ and $TF \perp FB$ (tan $\bot$ radius)
We have already proven above that $TC \perp CB$ and $AF \perp FB$ ; this option just repeats that fact (with the wrong reason). We still need to explain why this means that $T$ is the centre.
C ✔	$T$ is the point where two diameters meet.
The point where two diameters intersect is the centre of the circle.
D ✘	The diameters are perpendicular to the tangents (converse tan $\bot$ diameter).
We have already proven above that $AF$ is a diameter and $TC$ lies on a diameter; this option just repeats that fact. We still need to explain why this means that $T$ is the centre.

Statement	Reason
$\angle E_{2} = \angle F_{2}$
Also, $\angle F_{2} = \angle A_{2}$
$\therefore$
$\therefore TE = TA$

A ✘	$T$ is a point the same distance from $E$ and $A$ on the circle.
We can't be sure $T$ is the centre if we only know $TA = TE$ . There are infinitely many points in a circle that are the same distance from two points on the circumference. In the diagram below, $D$ is the same distance from $A$ and $E$ , but it is not the centre of the circle. However, if a point is the same distance from at least three points on the circumference, then that point is the centre of the circle.
B ✔	$TC = TA = TE$
This means that $T$ is a point the same distance from $C$ , $E$ and $A$ on the circle. So, $T$ is the same distance from three points on the circumference.
C ✘	$TA \perp AF$ and $TE \perp EF$ (tan $\bot$ radius)
This option uses the fact that $T$ is the centre of the circle. But, this is what we are trying to prove.
D ✘	$TC$ and $TA$ are radii (converse tan $\bot$ radius)
The reason converse tan $\bot$ radius can be used to prove that a line is part of a radius. To do so, we need to prove that the line is perpendicular to a tangent. But we have not done that in this proof.

Statement	Reason
$\angle E_{3} = \angle T_{3}$
$\therefore$ $= \text{90}\text{°}$
$\therefore$ $= \text{90}\text{°}$